Five numbers are in A.P., whose sum is 25

Let 5 terms of A.P. be

$a-2 d, a-d, a, a+d, a+2 d$

Sum $=25 \Rightarrow 5 a=25 \Rightarrow a=5$

Product $=2520$

$(5-2 d)(5-d) 5(5+d)(5+2 d)=2520$

$\Rightarrow \quad\left(25-4 d^{2}\right)\left(25-d^{2}\right)=504$

$\Rightarrow \quad 625-100 d^{2}-25 d^{2}+4 d^{4}=504$

$\Rightarrow 4 d^{4}-125 d^{2}+625-504=0$

$\Rightarrow 4 d^{4}-125 d^{2}+121=0$

$\Rightarrow 4 d^{4}-121 d^{2}-4 d^{2}+121=0$

$\Rightarrow\left(d^{2}-1\right)\left(4 d^{2}-121\right)=0$

$\Rightarrow \quad d=\pm 1, d=\pm \frac{11}{2}$

$d=\pm 1$ and $d=-\frac{11}{2}$, does not give $\frac{-1}{2}$ as a term

$\therefore \quad d=\frac{11}{2}$

$\therefore \quad$ Largest term $=5+2 d=5+11=16$